\(\int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [332]

Optimal result

Integrand size = 27, antiderivative size = 123 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \]

[Out]

-2*a^(3/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d+2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/5*a^
2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+2/5*a*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2953, 3055, 3060, 2852, 212} \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {2 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 d} \]

[In]

Int[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x])/(5*d*Sqrt[a + a
*Sin[c + d*x]]) + (2*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(5*d) + (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/
2))/(5*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2953

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 3055

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{5/2} \, dx}{a^2} \\ & = \frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {2 \int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \left (\frac {5 a^2}{2}-\frac {3}{2} a^2 \sin (c+d x)\right ) \, dx}{5 a^2} \\ & = \frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+\frac {4 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {15 a^3}{4}+\frac {3}{4} a^3 \sin (c+d x)\right ) \, dx}{15 a^2} \\ & = -\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}+a \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d} \\ & = -\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 a \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.42 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.18 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {(a (1+\sin (c+d x)))^{3/2} \left (5 \cos \left (\frac {3}{2} (c+d x)\right )-\cos \left (\frac {5}{2} (c+d x)\right )-10 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+10 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )\right )}{10 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

((a*(1 + Sin[c + d*x]))^(3/2)*(5*Cos[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 10*Log[1 + Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 10*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x)
)/2]))/(10*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (5 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )-\left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}+5 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-5 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{5 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(123\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(5*a^(5/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))-(a-a*sin(d*x+
c))^(5/2)+5*a*(a-a*sin(d*x+c))^(3/2)-5*a^2*(a-a*sin(d*x+c))^(1/2))/a/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (105) = 210\).

Time = 0.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.29 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {5 \, {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right )^{2} + 3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{10 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/10*(5*(a*cos(d*x + c) + a*sin(d*x + c) + a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x
+ c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x
+ c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x
+ c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) -
(a*cos(d*x + c)^2 + 3*a*cos(d*x + c) + a)*sin(d*x + c) + a)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) + d*sin
(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \csc \left (d x + c\right ) \,d x } \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2*csc(d*x + c), x)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.34 \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (16 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \sqrt {2} a \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 20 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{10 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/10*sqrt(2)*(16*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 40*a*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 5*sqrt(2)*a*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*
d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 20*a*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) \cot (c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\sin \left (c+d\,x\right )} \,d x \]

[In]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^(3/2))/sin(c + d*x),x)

[Out]

int((cos(c + d*x)^2*(a + a*sin(c + d*x))^(3/2))/sin(c + d*x), x)